Graphing a Vector Function

Discuss the graph of the vector function \[ \begin{equation*} \mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}\qquad 0\leq t\leq 2\pi \end{equation*}

Solution The components of \(\mathbf{r}=\mathbf{r}(t)\) are the parametric equations \begin{equation*} x=x(t) =\cos t\qquad y=y(t) =\sin t\qquad 0\leq t\leq 2\pi \end{equation*}

Since \(x^{2}+y^{2}=\cos ^{2}t\,{+}\,\sin ^{2}t=1\), the vector function \(\mathbf{r=r}(t)\) traces out a circle with its center at the origin and radius \(1\). See Figure 4. The curve begins at \(\mathbf{r}( 0) =\cos 0\,\mathbf{i}\,{+}\,\sin 0\,\mathbf{j}=\mathbf{i},\) contains the vectors \(\mathbf{r} \!\left( \dfrac{\pi }{2}\right) =\cos \dfrac{\pi }{2}\,\mathbf{i}\,{+}\,\sin \dfrac{\pi }{2}\,\mathbf{j}=\mathbf{j},\) \(\mathbf{r}(\pi) =\cos \pi \mathbf{i}\,{+}\,\sin \pi \mathbf{j}=-\mathbf{i},\) and \(\mathbf{r}\!\left( \dfrac{3\pi }{2}\right) =-\mathbf{j}.\) It ends at \(\mathbf{r}( 2\pi )=\mathbf{i},\) so the positive direction of the circle is counterclockwise.